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ilikephysics2 Group Title

Find the first 4 terms of the taylor series of sin x at x = pi/4.

  • 2 years ago
  • 2 years ago

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  1. ilikephysics2 Group Title
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    The first 4 derivatives are cos(x) then -sin(x) then -cos(x) and then sin(x)

    • 2 years ago
  2. Algebraic! Group Title
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    sin(pi/4) + cos(pi/4) * (x-pi/4) -sin(pi/4) * (1/2!) (x-pi/4)^2 .....

    • 2 years ago
  3. ilikephysics2 Group Title
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    is that it?

    • 2 years ago
  4. Algebraic! Group Title
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    pretty straightforward.. what was your question about?

    • 2 years ago
  5. Algebraic! Group Title
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    that's the first 3 terms... you should be able to get the 4th now?

    • 2 years ago
  6. ilikephysics2 Group Title
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    That does not look right though

    • 2 years ago
  7. Algebraic! Group Title
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    well... I lied a bit... a few of those terms are zero...

    • 2 years ago
  8. Algebraic! Group Title
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    but you can work from them... right?

    • 2 years ago
  9. RadEn Group Title
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    f(x) = f(x0) + (x-x0)*f'(x0)/1! + (x-x0)^2*f"(x0)/2! + (x-x0)^3*f'"(x0)/3!

    • 2 years ago
  10. RadEn Group Title
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    with x0 = pi/4, for f(x) = sinx --> f(pi/4) = sin(pi/4) = 1/2*sqrt(2) f'(x) = cosx --> f'(pi/4) = cos(pi/4) = 1/2*sqrt(2) f"(x) = -sinx --> f"(pi/4) = -sin(pi/4) = -1/2*sqrt(2) f'"(x) = -cosx --> f'"(pi/4) = -cos(pi/4) = -1/2*sqrt(2) now, just subtitute all numbers above to generally formula !!!

    • 2 years ago
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